Question

физика 8 н16эф С высоты девятиэтажного здания упал молоток массой 900 г на железную пластину массой 3 кг и остановился. На сколько градусов нагрелась пластина, если на её нагревание израсходовалось 25 % выделившегося при ударе количества теплоты? Высоту этажа принять равной 3 м.

Answer 1

Ответ: 0,04 градусов.

Объяснение: m*g*H = Q 0,25*Q = c*M*ΔT ΔT = 0,25*m*g*H/c*M = 0,25*0,9*10*27/460*3 = 0,04 градусов.

Answer 2

Problem Analysis

To solve this problem, we can use the principles of conservation of mechanical energy and the relationship between work, energy, and heat. The potential energy of the hammer at the initial height will be converted into kinetic energy as it falls and then into heat upon impact with the iron plate. We can use the work-energy principle to find the heat generated and then use the specific heat capacity to calculate the temperature rise of the iron plate.

Calculation Steps

1. Calculate the potential energy of the hammer at the initial height. 2. Use the work-energy principle to find the heat generated upon impact. 3. Calculate the temperature rise of the iron plate using the specific heat capacity.

Calculation

1. Calculate the potential energy of the hammer at the initial height: - The mass of the hammer, m = 900 g = 0.9 kg - The height of the building, h = 9 floors * 3 m/floor = 27 m - The acceleration due to gravity, g = 9.81 m/s^2 - The potential energy, PE = mgh - Substituting the values, PE = 0.9 kg * 9.81 m/s^2 * 27 m = 23.7153 J

2. Use the work-energy principle to find the heat generated upon impact: - The work done by the hammer upon impact is equal to the change in kinetic energy, which is equal to the initial potential energy. - The heat generated, Q = PE = 23.7153 J - Given that 25% of the heat is used to raise the temperature of the iron plate, the heat used for heating, Q_heat = 0.25 * Q = 0.25 * 23.7153 J = 5.928825 J

3. Calculate the temperature rise of the iron plate using the specific heat capacity: - The mass of the iron plate, m_plate = 3 kg - The specific heat capacity of iron, c_iron = 450 J/kg°C - The temperature rise, ΔT = Q_heat / (m_plate * c_iron) - Substituting the values, ΔT = 5.928825 J / (3 kg * 450 J/kg°C) ≈ 0.0044°C

Conclusion

The iron plate would have heated up by approximately 0.0044°C due to the impact of the hammer from the height of the nine-story building.